MAT 444 Test 4 Instructor: Helene Barcelo April 19, 004 Name: You can take up to hours for completing this exam. Close book, notes and calculator. Do not use your own scratch paper. Write each solution in the space provided, not on scratch paper. If you need more room, write on the back of the page. If you still need more room, ask for scratch paper. Show your reasoning on all problems; do not simply write an answer. Your solutions must be complete and organized, otherwise points may be deducted. Do all 6 problems. There are a total of 75 points.
Test 4 Page 1. (10 points) True-False. Give a brief reason (no detailed proofs) for each answer. (a) is the field of fractions of. The field of fraction of is. (b) is the field of fractions of. is its own field of fraction. (c) x 6 is irreducible over. x 6= ( x ) and is a unit in. (d) A polynomial f(x) of degree n with coefficients in a field F can have at most n zeros in F. Since F is a field, α F is a root ( x α) is a factor of f ( x ). So if degree f ( x) = n there are at most n roots since there are no zero divisors in F. (e) A polynomial f ( x ) of degree n with coefficients in a field F can have at most n zeros in any given field E such that F E. A similar argument hold, α E is a root ( x a) divides f ( x ). (f) is an ideal in. { ra r a } (g) Every ideal in a ring is a subring of the ring. If we do not require to have a 1 in the definition of a ring. If we use Artin s definition of a ring.
Test 4 Page (h) If D is an integral domain, then D[x] is an integral domain. Since D is an integral domain pxqx ( ) ( ) = 0 px ( ) 0 or qx ( ) 0, since deg p( xqx ) ( ) = deg px ( ) + deg qx ( ). (i) If R is any ring and f ( x) and g( x) in R[ x ] are of degrees and 4, respectively, then f ( xgx ) ( ) is always of degree 7. Consider 6[ x]. Then 4 x x 0 but x 0 and 4 x 0 in 7[ x]. (j) If R is a ring, then x is never a zero divisor in R[ x ]. True R[ x ] can be viewed as adjoining an element to R with only the relation that 0.x = 0.
Test 4 Page 4. An ideal P R in a commutative ring R with unity is a prime ideal if ab P implies that either a P of b P for ab, R. (a) (5 points) Show that {0} is a prime ideal of. Done in class. (b) (5 points) Let P R be an ideal of a commutative ring R with unity. Prove that R / P is an integral domain if and only if P is a prime ideal. Done in class.
Test 4 Page 5. (10 points) Let R = [ α] be the ring obtained from by adjoining an element α that satisfies the relation a = 0, and let S = [ β ] be the ring obtained from by joining an element β that satisfies the relation β = 0. (a) Describe both rings R and S. [ α] [ ] and [ β] [ ] (b) Prove that R and S are not isomorphic rings. We have that R [ x]/( x ) and S [ x]/( x ) If R S then any isomorphism ϕ : R S is such that ϕ (1 R) = 1 s. But 1R = 1S = 1, thus ϕ ( n) = n n. But = ϕ( ) ϕ( ) = where ϕ ( ) = a + b a, b. ( ) a + b = a + ab + b = a + b = and ab = 0 a = 0 or b = 0. But this is impossible, for if a = 0 b = which has no solution in,if b = 0then a = which does not have any solution in either.
Test 4 Page 6 4. Let R = [ 6]. (a) (5 points) Prove that the only units of R are ± 1. a + b 6 is a unit 6 6 1 a + b = a + b = ± i.e. if b = 0 and a =± 1. (b) (5 points) Prove that R is not a unique factorization domain by finding two different factorizations of 10. 10 =.5 = ( + 6 )( 6 ). Note that no elements of [ 6] have length 5, since a + 6b = 5 is clearly impossible. Hence, + 6 and 6 are irreducible, since each have length 10 and any decomposition would involve an element of length 5. Similarly, 5 has length 5 and is irreducible. Moreover, a + 6b = is impossible, so no element can have length, and hence (of length 4) is irreducible. (c) (5 points) Find an example of an irreducible element that is not prime. + 6 is irreducible but not prime, for it divides 10 but does not divide either or 5.
Test 4 Page 7 5. (5 points) Find all c such that is a field. [ x]/( x + cx + 1) [ x]/( x + cx + 1) is a field irreducible polynomial. ( x + cx + 1) is a maximal ideal x + cx + 1 is an if c = 0 then x + 1 has a root x = 1 = mod thus x + 1 is reducible over if c = 1 x + x + 1 has a root x = 1, so it is reducible over if c = x + x + 1 0 x = 0,1, hence x + x + 1 is irreducible over. Hence for all c = (mod), [ x]/( x + cx + 1) is a field.
Test 4 Page 8 6. (5 points) Show that the characteristic of an integral domain D must be either 0 or a prime p. (Hint: what would happen if the characteristic of D was mn? Consider n(1 D) = 1D + + 1D D. If n(1 D ) 0 n then characteristic of D = 0. Otherwise, there exist n such that n(1 D ) = 0. Let n 0 be the smallest such n. Claim n 0 is a prime integer. Suppose not, then n0 = km, m < n0 k, m > 1 km(1 D) = 0 (1D + + 1 D) (1D + + 1 D) = 0 k times m times Let (1 + + 1 ) = a and (1 + + 1 ) = b D D D D k times m times Since D is an integral domain, then ab = 0 a = 0 or b = 0. In either case this violates the minimality of n 0.